Question

2. Concentration: A solution preparation guide calls for the use of 37.50 g KH2PO4 with 489.4 g water. The volume of the solution is 500.0 mL. (A) Determine the mass percentage of KH2PO in this solution. [B] Determine the mass-volume percentage of KH2PO4 in this solution. [C] Determine the molarity of KH2PO4 in this solution.

Answer 1

A)
Mass % = mass of KH2PO4 * 100 / mass of solution
= 37.50 * 100 / (37.50 + 489.4)
= 7.117 %
Answer: 7.117 %

B)
(m/v)% = mass of KH2PO4 * 100 / volume of solution
= 37.50 * 100 / 500.0
= 7.500 %
Answer: 7.500 %

C)

Molar mass of KH2PO4,
MM = 1*MM(K) + 2*MM(H) + 1*MM(P) + 4*MM(O)
= 1*39.1 + 2*1.008 + 1*30.97 + 4*16.0
= 136.086 g/mol


mass(KH2PO4)= 37.50 g

use:
number of mol of KH2PO4,
n = mass of KH2PO4/molar mass of KH2PO4
=(37.5 g)/(1.361*10^2 g/mol)
= 0.2756 mol
volume , V = 5*10^2 mL
= 0.5 L


use:
Molarity,
M = number of mol / volume in L
= 0.2756/0.5
= 0.5511 M
Answer: 0.5511 M